Midterm 1 Solutions
1.
(a) Find 2
×
2 matrices
A
and
B
such that
AB
6
=
BA
.
There are many examples to answer this problem. Here is one:
Let
A
=
1
0
0
0
and
B
=
0
1
1
1
.
Then
AB
=
0
1
0
0
and
BA
=
0
0
1
0
so that
AB
6
=
BA
.
(b) For general
n
×
n
matrices
A
and
B
find an expression for (
A
+
B
)
2
.
Let
A
and
B
be
n
×
n
matrices. Then,
(
A
+
B
)
2
= (
A
+
B
)(
A
+
B
) =
A
(
A
+
B
) +
B
(
A
+
B
) =
A
2
+
AB
+
BA
+
B
2
.
Where the last two equalities are satisfied by distributive properties of matrix operations.
(c) Write the following system of Differential Equations in matrix vector form:
x
0
1
= 3
x
1
+ 5
x
2
+ sin(
t
)
,
x
0
2
=
x
1
+ 2
x
2
+ cos(
t
)
x
0
1
x
0
2
=
3
5
1
2
x
1
x
2
+
sin(
t
)
cos(
t
)
2. Answers without
correct
supporting work will not count, even if correct. Let
A
=
1

1
2
2
1
11
4

3
10
.
Compute
A

1
using Elementary Row Transformation and check your work by computing
A

1
A
.
1

1
2

1
0
0
2
1
11

0
1
0
4

3
10

0
0
1
→
1
1

1
2

1
0
0
0
3
7


2
1
0
0
1
2


4
0
1
→
2
1

1
2

1
0
0
0
1
2


4
0
1
0
3
7


2
1
0
→
3
1
0
4


3
0
1
0
1
2


4
0
1
0
0
1

10
1

3
→
4
1
0
0


43

4
13
0
1
0


24

2
7
0
0
1

10
1

3
.
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 Spring '07
 Guralnick
 Math, Determinant, Matrices, Characteristic polynomial, Diagonal matrix, Triangular matrix, Upper Triangular Matrix

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